SOLUTION: how to prove trigonometry identity proof: 1+sinx-cosx/1+sinx+cosx = sin2x(1-cosx)/cosx(1-cos2x)

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Question 612387: how to prove trigonometry identity proof:
1+sinx-cosx/1+sinx+cosx = sin2x(1-cosx)/cosx(1-cos2x)
Found 2 solutions by jsmallt9, Edwin McCravy:
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Please re-post and put each numerator and each denominator in parentheses. What you posted means:
1%2Bsinx-cosx%2F1%2Bsinx%2Bcosx+=+sin2x%281-cosx%29%2Fcosx%281-cos2x%29
which I doubt is the actual equation.

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
He's right about your incomplete use of parentheses.  You should have

(1+sinx-cosx)/(1+sinx+cosx) = (sin2x(1-cosx))/(cosx(1-cos2x))

If you are writing an expression all on one line, then if a numerator or
denominator contains more than one letter or number, you must put a "(" 
to show where the numerator or denominator begins and a ")" to show where
it ends. I will now prove what you should have.

Since both sides are complicated, this is one in which we will have to work 
each side separately, simplifying each to the same expression.  
First we simplify the left side:

%281%2Bsinx-cosx%29%2F%281%2Bsinx%2Bcosx%29 = sin2x%281-cosx%29%2Fcosx%281-cos2x%29

%281%2B%28sinx-cosx%29%29%2F%281%2B%28sinx%2Bcosx%29%29

To make things easier use S = sin(x) and C = cos(x) on the left side:

%281%2B%28S-C%29%29%2F%281%2B%28S%2BC%29%29

Multiply by %281-%28S%2BC%29%29%2F%281-%28S%2BC%29%29 which just equals 1

%281%2B%28S-C%29%29%2F%281%2B%28S%2BC%29%29·%281-%28S%2BC%29%29%2F%281-%28S%2BC%29%29

%281-%28S%2BC%29%2B%28S-C%29-%28S-C%29%28S%2BC%29%29%2F%281-%28S%2BC%29%5E2%29

%281-S-C%2BS-C-%28S%5E2-C%5E2%29%29%2F%281-%28S%5E2%2B2SC%2BC%5E2%29%29

%281-2C-S%5E2%2BC%5E2%29%2F%281-S%5E2-2SC-C%5E2%29%29

%28%281-S%5E2%29-2C%2BC%5E2%29%2F%28%281-S%5E2%29-2SC-C%5E2%29%29

Use 1-Sē=Cē which is really 1-sin%5E2%28x%29=cos%5E2%28x%29

%28C%5E2-2C%2BC%5E2%29%2F%28C%5E2-2SC-C%5E2%29%29

%282C%5E2-2C%29%2F%28-2SC%29%29

%282C%28C-1%29%29%2F%28-2SC%29

%28cross%282%29cross%28C%29%28C-1%29%29%2F%28-cross%282%29S%2Across%28C%29%29

%28%28C-1%29%29%2F%28-S%29

Multiply by %28-1%29%2F%28-1%29

%28-C%2B1%29%2FS

%281-C%29%2FS

%281-cos%28x%29%29%2Fsin%28x%29

Now we simplify the right side:

%28sin%282x%29%281-cosx%29%29%2F%28cosx%281-cos2x%29%29

We use sin(2x)=2sin(x)cos(x) and
cos(2x)=1-2sinē(x)

%282sin%28x%29cos%28x%29%281-cosx%29%29%2F%28cosx%281-%281-2sin%5E2%28x%29%29%29%29

%282sin%28x%29cos%28x%29%281-cosx%29%29%2F%28cosx%281-1%2B2sin%5E2%28x%29%29%29

%282sin%28x%29cos%28x%29%281-cosx%29%29%2F%28cosx%282sin%5E2%28x%29%29%29



%281-cos%28x%29%29%2Fsin%28x%29

So we have reduced both sides to the same expression, so we
have proved the identity.

Edwin